Rhy1399 wrote:
Every 5 couples that George matches, 3 fall in love. If he matches 4 couples what is the probability of at least one couple not falling in love?
I know the 1-(all get married) approach. But I dont get why I am getting the wrong answer with the individual scenario approach? Can anyone solve with that method? (1 does not fall in love+ 2 do not fall in love....)
P of a couple falling in love - \(\frac{3}{5}\)
P of a couple NOT falling in love - \(\frac{2}{5}\)
All falling in love = \(\frac{3}{5}^4=\frac{81}{625}\)
the probability of at least one couple not falling in love=1-P(all get married)=\(1-\frac{81}{625}=\frac{544}{625}\)
Individual P P of a couple falling in love - \(\frac{3}{5}\)
P of a couple NOT falling in love - \(\frac{2}{5}\)
Just one couple not falling in love = \(4C1*\frac{2}{5}*\frac{3}{5}^3=4*\frac{54}{625}=\frac{216}{625}\). May be you are missing out on 4C1 and getting your answer wrong.
Two couples not falling in love = \(4C2*\frac{3}{5}^2*\frac{2}{5}^2=6*\frac{36}{625}=\frac{216}{625}\)
Three couples not falling in love = \(4C3*\frac{3}{5}*\frac{2}{5}^3=4*\frac{24}{625}=\frac{96}{625}\)
All couples not falling in love = \(4C4*\frac{2}{5}^4=1*\frac{16}{625}=\frac{16}{625}\)
Total P = \(\frac{96+216+216+16}{625}=\frac{544}{625}\)